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$$\int_3^5\frac{t}{1+0.1t} dt $$

For some reason this is equal to:

1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))

I have no idea how to reduce to that.

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  • $\begingroup$ If you let $x=1+0.1t$, then $t=10x-10$... $\endgroup$ – Eleven-Eleven Feb 18 at 13:53
  • $\begingroup$ Do you mean $$\int_{3}^{5}\frac{t}{1+\frac{1}{10}t}dt$$? $\endgroup$ – Dr. Sonnhard Graubner Feb 18 at 13:54
  • $\begingroup$ This is not an improper integral, by the way. So, I removed that tag. $\endgroup$ – Michael Rybkin Feb 18 at 14:29
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Hint:

$$\frac{t}{1+0.1t} = \frac{10\cdot(1+0.1t) - 10}{1+0.1t} = 10 - \frac{10}{1+0.1t}$$

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$$ \frac{x}{1+0.1x}=\frac{x}{1+0.1x}\cdot\frac{10}{10}= \frac{10x}{10+x}=10\left(\frac{x}{10+x}\right)=\\ 10\left(\frac{-10+10+x}{10+x}\right)= 10\left(\frac{-10}{10+x}+\frac{10+x}{10+x}\right)= 10\left(-\frac{10}{10+x}+1\right)=\\ 10\left(1-\frac{10}{10+x}\right)=10-\frac{100}{10+x}. $$


$$ \int\left(10-\frac{100}{10+x}\right)\,dx= 10\int\,dx-100\int\frac{1}{10+x}\frac{d}{dx}(10+x)\,dx=\\ 10x-100\int\frac{1}{10+x}\,d(10+x)= 10x-100\ln{|10+x|}+C. $$
$$ \int_3^5\frac{t}{1+0.1t}\,dt= \bigg[10t-100\ln{|10+t|}\bigg]_3^5=\\ 50-100\ln{15}-(30-100\ln{13})= 20-100\ln{15}+100\ln{13}=\\ 20-100(\ln{15}-\ln{13})=20-100\ln{\frac{15}{13}}. $$
The answer you gave is equivalent to what I got: $$ \frac{1}{0.1}\left(2-\frac{1}{0.1}\left[\ln{1.5}-\ln{1.3}\right]\right)= 10\left(2-10\left[\ln{\frac{15}{10}}-\ln{\frac{13}{10}}\right]\right)=\\ 20-100\ln{\left(\frac{15}{10}\div\frac{13}{10}\right)}= 20-100\ln{\frac{15}{13}}. $$

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